Physical Sciences Grade 12 KZN April 2021 Memo

Downloaded from hlayiso.com } KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE GRADE 12 PHYSICAL SCIENCES COMMON TEST APRIL 2021 MARKS : 100 TIME : 2 Hours This question paper consists of 11 pages and 2 data sheets Please turn over Copyright reserved

Downloaded from hlayiso.com 2 Physical Sciences NSC-Grade12 April 2021 Common Test INSTRUCTIONS AND INFORMATION A Write your centre number and examination number in the appropriate spaces on the ANSWER BOOK. 2 This question paper consists of SEVEN questions. Answer ALL the questions in the ANSWER BOOK. . Start EACH question on a NEW page in the ANSWER BOOK. 4. Number the answers correctly according to the numbering system used in this question paper. 5. Leave ONE line between two sub-questions, for example between QUESTION 2.1 and QUESTION 2.2. 6. You may use a non-programmable calculator. ie You may use appropriate mathematical instruments. 8. You are advised to use the attached DATA SHEETS. 9. Show ALL formulae and substitutions in ALL calculations. 10. Round off your final numerical answers to a minimum of TWO decimal places. 11. Give brief motivations, discussions et cetera where required. 2: Write neatly and legibly. Copyright reserved Please turn over

Downloaded from hlayiso.com 3 Physical Sciences NSC-Grade12 April 2021 Common Test QUESTION 1: MULTIPLE-CHOICE QUESTIONS Various options are provided as possible answers to the following questions. Choose the answer and write only the letter (A-D) next to the question number (1.1-1.6) in the ANSWER BOOK, for example 1.11 D. 1.1. Acar of mass 1000 kg pulls a trailer of mass 500 kg as shown. - ae i eee, el According to Newton’s third law of motion, which ONE of the following statements is TRUE? A The car and trailer pull each other with a force that is equal in magnitude but opposite in direction. Therefore, the net force is zero and the trailer cannot move. B The force that the car exerts on the trailer is greater than the force that the trailer exerts on the car. Therefore, the trailer moves forward. Cc The action force from the car is quicker than the reaction force from the trailer, so they move forward. D The action-reaction forces are equal in magnitude, but the force between the ground and wheels pushes them forward. (2) 1.2 When the momentum of an object of constant mass is doubled, then its kinetic energy will be ... A halved. B doubled. ~-- 5+ ee C - -~threetimes greater- - - - - - ----- Hee Sid D four times greater. (2) 1.3. Two objects are released from the same height at the same time. One object has TWICE the weight of the other. Neglecting friction, which ONE of the following statements is CORRECT for the motion? A The heavier object hits the ground first B The lighter object hits the ground first. Cc Both objects hit the ground at the same time D On hitting the ground, the heavier object has a greater velocity than the lighter object (2) Copyright reserved Please turn over

Downloaded from hlayiso.com 4 Physical Sciences NSC-Grade12 April 2021 Common Test 1.4 Which ONE of the following compounds is a KETONE? A ~~ CH3COCH2CH3 B = CHsCOOCH2CH3 C CH3CH2zCH2CHO D CH3sCH(OH)CH2CH3 (2) 1.5 P and Q represent two organic compounds in the reactions below: REACTION I: ethene 48", p NaOH(aq) _—_—— heat REACTION II: P Q Which ONE of the following represents compound Q? A CH2CH2 B CH3CH3 C CH3CH20H D CH3CH2Br (2) 1.6 CnH2nO2 is the general formula for both... A Aketone and an aldehyde. B- Anester and an aldehyde. C Aketone and a carboxylic acid. D Anester and a carboxylic acid. (2) [12] *~ Copyright reserved : Please turn over

Downloaded from hlayiso.com 5 Physical Sciences NSC-Grade12 QUESTION 2 2.1. A box of mass of 50 kg, placed on a ROUGH horizontal surface, April 2021 Common Test is pulled by a force of 120 N that acts at an angle of 60° to the horizontal. The box moves at a CONSTANT VELOCITY along the surface. 2.1.1. State Newton's First Law of Motion in words. moves. forces acting on the box? 2.1.4 Calculate the normal force acting on the box. magnitude of the normal force? EXPLAIN the answer. 2.2. The Sun and the Earth exert a gravitational force on each other. Sun 2.2.1. State Newton's law of Universal Gravitation in words. 2.2.2. The mass of the Sun is 332 600 times greater than that (2) 2.1.2. Draw a free-body diagram of ALL the forces acting on the box while it (4) 2.1.3 The box is travelling at constant velocity. What can be deduced about the (1) (3) 2.1.5 The angle of the applied force is now decreased. How will this affect the Choose from INCREASES, DECREASES or REMAINS THE SAME (3) (2) of the Earth. qhe distance between the centres of the Sun and the Earth is 1,496 x 10 m. Calculate the gravitational force that the Sun exerts on the Earth. (4) [19] Copyright reserved Please turn over

Downloaded from hlayiso.com 6 Physical Sciences NSC-Grade12 April 2021 Common Test QUESTION 3 A goods train is being assembled in a yard. Car 2, of mass of 92 x 10° kg and moving with a velocity of 1.3 mss” to the right collides with Car 1 of mass 65 x 10° kg that is moving to the right with a velocity of 0.80 mes. The two cars are joined together after the collision. Ignore the effects of friction. 1.3 mes" 0,8 mss" v - - : ; Before collision : After collision 3.1. State the principle of conservation of momentum in words (2) 3.2 Calculate the magnitude of the velocity of the cars after the collision. (4) 3.3. Determine, by means of appropriate calculations, whether the collision between the two cars is elastic or inelastic. (5) [11] QUESTION 4 A hot air balloon is rising vertically at a constant velocity of 15 mes’'. When the balloon is at some unknown height above the ground, a stone is released from the balloon. The stone is observed to hit the ground with a velocity of 45 mss”. 4.1. Define free fall (2) 4.2 Write down the magnitude and direction of the initial velocity of the stone when it is released (2) Copyright reserved Please turn over

Downloaded from hlayiso.com 7 Physical Sciences NSC-Grade12 April 2021 Common Test 4.3. Determine how high above the ground the hot air balloon was when the stone was released from it. (3) 4.4 Determine how long the stone took to hit the ground after being released. (3) 4.5 Sketch the position-time graph for the entire motion of the stone from the moment it ' was released from the hot air balloon to the moment it hits the ground. Use the ground as a zero reference point. Indicate the following on your graph: e Height when the stone is released. e Time taken to hit the ground. (4) [14] Copyright reserved Please turn over

Downloaded from hlayiso.com 8 Physical Sciences NSC-Grade12 QUESTION 5 (Start on a new page.) The letters A to G in the table below represent organic compounds. * | H —C—H roo | Ho ¢ Cc Cc c¢ ¢ | | | | H H H H a HH ¢ F | H ; 1. Pf ce ace ien ee e H H H : Ff fl i. Ho CG ec Ld | | H H H — C —H | n T D | CyHg0 E | CHsCH(CHs)CH(CHs)CH2CCCHs | F | CHs(CHz)sCH2OH Use the above table to answer the following questions: 5.1. Write down the: 5.1.1. IUPAC name of compound A. 5.1.2 IUPAC name of compound C. 5.1.3. Name of the homologous series to which compound B belongs. Copyright reserved Please turn over April 2021 Common Test (2) (2) (1)

Downloaded from hlayiso.com 9 Physical Sciences NSC-Grade12 April 2021 Common Test 5.2 Compound E is a hydrocarbon. 5.2.1 Define the term hydrocarbon. 5.2.2 Is compound E a saturated or unsaturated compound? Give a reason for the answer. 5.2.3. Write down the general formula of the homologous series to which compound E belongs. 5.3. Write down the IUPAC name of a POSITIONAL ISOMER of compound F. 5.4 Dis the molecular formula of TWO functional isomers. 5.4.1. Define the term functional isomer. 5.4.2 Draw the STRUCTURAL FORMULA of ONE of the FUNCTIONAL isomers of D. 5.4.3. Write down the IUPAC name of the OTHER FUNCTIONAL isomer of D. (Hint: This is NOT the same compound as mentioned in Question 5.4.2) (2) (1) (2) (2) (2) [18] Copyright reserved Please turn over

Downloaded from hlayiso.com 10 Physical Sciences NSC-Grade12 April 2021 Common Test QUESTION 6 (Start on a new page.) The boiling points of TWO organic compounds, A and B, were determined. FORMULA MOLECULAR MASS (g.mo!") A CH3(CH2)2COOH 88,1 B CH3(CH2)3CH20H 88,1 6.1. State the definition of boiling point. (2) The following boiling points were obtained: 137 °C 163 °C 6.2 Write down the boiling point that is most likely to be that of compound A. (2) 6.3 Explain FULLY how you arrived at the answer. (4) 6.4 Write down the IUPAC name of the compound with the LOWER vapour pressure. (2) 6.5 The boiling point of another compound, CH3(CH2)2CH20H was also determined. How will the boiling point of this compound compare to that of compound B? Choose from GREATER THAN, LESS THAN or EQUAL TO. Explain the answer. (3) [13] Copyright reserved Please turn over

Physical Sciences Downloaded from hlayiso.com 11 NSC-Grade12 April 2021 Common Test QUESTION 7 (Start on a new page.) The flow diagram shows how COMPOUND X can be used to prepare other organic compounds. The letters A, B, C and D represent different organic reactions. Compound X is a HALOALKANE. CHsCOOCH2CH2CH3 COMPOUND X Cc A B D . CH3CHOHCH3 ALKENE PRIMARY ALCOHOL (major product) Use the information in the flow diagram to answer the following questions: 7.1 Write down the type of: 7.1.1 Substitution reaction represented by A. (1) 7.1.2 Elimination reaction represented by C. (1) 7.2 Write down the type of reaction represented by: 724 Be (1) C2522 “Di (1) 7.3 Consider REACTION GC. Write down 7.3.1 TWO reaction conditions for this reaction. (2) 7.3.2 |\UPAC name of the alkene formed. (1) 7.4 Write down the name or formula of the inorganic reactant for REACTION A. (1) 7.5 Reaction B involves the reaction of an organic compound with a PRIMARY ALCOHOL. 7.5.1 Define the term primary alcohol. (2) 7.5.2 Write down the name or formula of the catalyst used in REACTION B. (1) 7.5.3 Write down the structural formula of the ORGANIC COMPOUND that reacts with the primary alcohol in REACTION B. (2) [13] TOTAL MARKS: [100] Copyright reserved Please turn over

Downloaded f from hlayiso. com ene erm eres cee — ()) 1 H 1 3 Li 7 1 ; Na 23 TABLE 3: THE PERIODIC TABLE OF ELEMENTS 5 6 7, 8 9 40 1 12) 13 14' °15 #16 «17 | 48 am = (iy) (vil) (vil) (vit) Atomic number KEY/SLEUTEL Atoomgetal He 29 Electronegativity ® Cu Symbol Elekt iwit = Si f lektronegatiwiteit_y| 63,5 |e imboo Approximate relative atomic mass Benaderde relatiewe atoommassa 19 K 39 23/ 24] 25| 26| 27) 28 ij2 V (2 Crj2Mn|2 Fe |2 Co/2 Ni |Z 51| 52; 55| 56| 59| 59 37 Rb 86 “| 42/43] 44! 457) 46 Nb |2 Mo|2 Te |S Ru|a Rh |S Pd 92 | 96 101| 103| 106 55 Cs 73 74 75 76 7 78 Ta} W/ Rej Os| Ir; Pt 181| 184] 186] 190 195 58 | 59 | 60 ] 61 63 Ce | Pr | Nd Eu 144 90 | 91 | 92 Th U 313 NVA 1498V 1 AMFgOle2d Fld -€ TAGVL SLN3AW3A14 dO 318VL DIGOlMAd 3HL ‘€ ATEVL ZL @PEIQ-OSN seouains jealskud ysa| UOWWOD 1Z0Z Idy

Downloaded from hlayiso.com 2 Physical Sciences NSC-Grade12 April 2021 Common Test DATA FOR PHYSICAL SCIENCES GRADE 12 PAPER 1 (PHYSICS) GEGEWENS VIR FIS/ESE WETENSKAPPE GRAAD 12 VRAESTEL 1 (FIS/KA) TABLE 1: PHYSICAL CONSTANTS/ TABEL 1: FISIESE KONSTANTES NAME/NAAM SYMBOL/ SIMBOOL VALUE / WAARDE Acceleration due to gravity g 9,8 m-s -2 Swaartekragversnelling Universal gravitational constant 6,67 x 10- 11 N·m2·kg-2 G Universe/e gravitasiekonstante Speed of light in a vacuum C 3 '0 x 108 m-s-1 Spoed van Jig in 'n vakuum Planck's constant h 6,63 X 10-34 J·S Planck se konstante Coulomb's constant k 9,0 X 109 N·m2 -c-2 Coulomb se konstante Charge on electron - e -1 ' 6 X 10-19 C Ladinq op electron Electron mass me 9, 11 X 10-31 kg Elektronmassa Mass of Earth M 5,98 X 10 24 kg Massa van Aarde Radius of Earth RE 6,38 X 106 m Radius van Aarde TABLE 2: FORMULAE/ TABEL 2: FORMULES MOTION/BEWEGING Vf =V j +a�t 2 1 1 /J.x = vi !J.t. + a�t or/of Liy =vi !J.t. + a�t 2 2 2 2 2 v 1 =V i +2aill<or/of v 1 =V i +2a�y tsx =(vi ;v 1 )Mor/of Liy =(vi ; v 1 )M FORCE/KRAG I Fnet = ma fslmax\ = IJ s N p=mv fk = IJkN Fnet ilt =4' W=mg 4:> = mv 1 -m v; Gm 1 m2 Gm F= 2 g=- r r2 WORK, ENERGY AND POWER/ARBEID, ENERGIE EN DRYWING W=Ffil<COS0 U= mgh or/ofE p = mgh Wnet = LiK or/of Wnet = LiEk 1 K=-mv If 2 or o E k 1 =-mv 2 2 2 Li K = K1- Ki or/of b.Ek = Ekf - Eki Wnc = �K + �U or/of Wnc =�Ek+ �E p P=W dt Pav = F·Vav / Pgem = F·Vgem Copyright reserved

Downloaded from hlayiso.com 1 | P a g e pppppppppppPhysical Sciences / P2 1 April 2021 Common Test Physical Sciences 1 April 2021 Common Test NSC-Guideline NATIONAL SENIOR CERTIFICATE GRADE 12 PHYSICAL SCIENCES COMMON TEST APRIL 2021 MARKING GUIDELINE MARKS : 100 TIME : 2 hours This Marking Guideline consists of 8 pages. Copyright Reserved Please Turn Over

Downloaded from hlayiso.com 2 | P a g e pppppppppppPhysical Sciences / P2 April 2021 Common Test 2 Physical Sciences 2 April 2021 Common Test NSC-Guideline QUESTION 1 1.1 D (2) 1.2 D (2) 1.3 C (2) 1.4 A  (2) 1.5 C (2) 1.6 D (2) [12] QUESTION 2 2.1.1 A body will remain in its state of rest or motion at constant velocity unless a non- zero resultant/net force acts on it.  (2) 2.1.2 N FA  FAy N fk  fk FAx W W CRITERIA FOR MARKING Acceptable Labels • Mark is awarded for label and arrow. N Normal/FN/FNormal • Do not penalise for length of arrows. fk Friction/Ff/fkinetic • Deduct 1 mark for any additional force. FA F/120 N/Fapplied • If force(s) do not make contact with body/dot W Fg/Fgravity/Gravitational force/490 N : Max:3/4 • If arrows missing but labels are there: Max:3/4 (4) 2.1.3 Sum of all forces acting on the box is zero/ No net force acting on the box (1) 2.1.4 Fnet = ma Fnet = 0  N+ FAy= W N + 120Sin 60o = 50(9,8)  N = 386,08 N (3) 2.1.5 INCREASES, If the angle of the applied force decreases, the vertical component of the applied force will also decrease. This will affect the normal force since the sum of the normal force and vertical component of the applied force must be equal to the weight of the object. (3) Copyright Reserved Please Turn Over

Downloaded from hlayiso.com 3 | P a g e pppppppppppPhysical Sciences / P2 3 April 2021 Common Test Physical Sciences 3 April 2021 Common Test NSC-Guideline 2.2.1 Each body in the universe attracts every other body with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.  (2) OR Each particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.  2.2.2    F = 3,55 1026 N (4) [19] QUESTION 3 3.1 The total linear momentum of a closed system remains constant (is conserved) OR The total linear momentum before collision is equal to the total linear momentum after the collision. (2) 3.2 Σpf =Σpi  (m1 + m2)v = m1vi1 + m2vi2 (92 x 103 + 65 x 103)v = 65 x 103 (0,8) + 92 x 103 (1,3) v = 1,09 m•s-1 ∴ the velocity of the cars is 1,09 m•s-1 (4) POSITIVE MARKING FROM Q. 3.2 3.3 ΣKf = ½ (m1 + m2)v2 ΣKi = ½ m1 vi12 + ½m2 vi22 = ½ (65 x 103 + 92 x 103) 1,09 2 = ½ (65 x 103 )(0,8)2 + ½(92 x 103)(1,3)2 = 93265,85 J = 98540 J ΣKf ≠ ΣKi ∴Collision is Inelastic  (5) [11] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com 4 | P a g e pppppppppppPhysical Sciences / P2 4 April 2021 Common Test Physical Sciences 4 April 2021 Common Test NSC-Guideline QUESTION 4 4.1 Motion upon which the only force acting is the force of gravity.  (2) 4.2 15 m•s-1 upwards (2) 4.3 StoneTrajectory OPTION 1:From A to D UPWARDS POSITIVE UPWARDS NEGATIVE vf2 = vi2 + 2aΔy vf2 = vi2 + 2aΔy -452 = 152 + 2(-9,8)Δy 452 = -152 + 2(9,8)Δy Δy = -91,84 m Δy = 91,84 m ∴ Height above the ground when the ∴ Height above the ground when the stone is released is 91,84 m stone is released is 91,84 m OPTION 2:From A to D UPWARDS POSITIVE UPWARDS NEGATIVE vf = vi + aΔt vf = vi + aΔt -45 = 15 +(-9,8) Δt 45 = -15 +(9,8) Δt Δt = 6,12 s Δt = 6,12 s Δy= viΔt + ½aΔt2 Δy= viΔt + ½aΔt2 Δy = (15)(6,12) + ½(-9,8)(6,12)2 Δy = (-15)(6,12) + ½(9,8)(6,12)2 Δy = -91,73 m Δy = 91,73 m ∴ Height above the ground when the ∴ Height above the ground when the stone is released is 91,73 m stone is released is 91,73 m OPTION 3:From A to D UPWARDS POSITIVE UPWARDS NEGATIVE vf = vi + aΔt vf = vi + aΔt -45 = 15 +(-9,8) Δt 45 = -15 +(9,8) Δt Δt = 6,12 s Δt = 6,12 s     Δy = -91,80 m Δy = -91,80 m ∴ Height above the ground when the ∴ Height above the ground when the stone is released is 91,80 m stone is released is 91,80 m Copyright Reserved Please Turn Over

Downloaded from hlayiso.com 5 | P a g e pppppppppppPhysical Sciences / P2 April 2021 Common Test 5 Physical Sciences 5 April 2021 Common Test NSC-Guideline OPTION 3:From C to D UPWARDS POSITIVE UPWARDS NEGATIVE vf = vi + aΔt vf = vi + aΔt -45 = -15 +(-9,8) Δt 45 = 15 +(9,8) Δt Δt = 3,06 s Δt = 3,06 s     Δy = -91,80 m Δy = 91,80 m ∴ Height above the ground when the ∴ Height above the ground when the stone is released is 91,80 m stone is released is 91,80 m Δy= viΔt + ½aΔt2 Δy= viΔt + ½aΔt2 Δy = (-15)(3,06) + ½(-9,8)(3,06)2 Δy = (15)(3,06) + ½(9,8)(3,06)2 Δy = -91,78 m Δy = 91,78 m ∴ Height above the ground when the ∴ Height above the ground when the stone is released is 91,78 m stone is released is 91,78 m (3) 4.4 OPTION 1:From A to D UPWARDS POSITIVE UPWARDS NEGATIVE vf = vi + aΔt vf = vi + aΔt -45 = 15 +(-9,8) Δt 45 = -15 +(9,8) Δt Δt = 6,12 s Δt = 6,12 s OPTION 2:From A to B and then C - D UPWARDS POSITIVE UPWARDS NEGATIVE vf = vi + aΔt vf = vi + aΔt 0 = 15 +(-9,8) Δt 0 = -15 +(9,8) Δt   Δt = 1,53 s Δt = 1,53 s vf = vi + aΔt vf = vi + aΔt -45 = -15 +(-9,8) Δt 45 = 15 +(9,8) Δt Δt = 3,06 s Δt = 3,06 s Δt = 2(1,53) + 3,06 Δt = 2(1,53) + 3,06 Δt = 6,12 s Δt = 6,12 s OPTION 3:POSITIVE MARKING FROM 7.2 From A to D UPWARDS POSITIVE UPWARDS NEGATIVE Δy= viΔt + ½aΔt2 Δy= viΔt + ½aΔt2 -91,73 = (15Δt+ ½(-9,8Δt2 91,73 = (-15Δt+ ½(9,8Δt2 Δt = 6,12 s Δt = 6,12 s (3) Copyright Reserved Please Turn Over

Downloaded from hlayiso.com 6 | P a g e pppppppppppPhysical Sciences / P2 6 April 2021 Common Test Physical Sciences 6 April 2021 Common Test NSC-Guideline 4.5 POSITIVE MARKING FROM 7.2 AND 7.3 91,80 Position/m 0 6,12 Time/s 0 OR Position/m Time/s 6,12 -91,80 NOTE: LEARNERS MAY CHOOSE HEIGHT WHEN THE BALL IS RELEASED AS THE REFERENCE POINT Criteria for marking Correct shape  Indication of height (9,73 – 9,84 m)  Indication of the correct end time  Correct choice of the zero point of reference  (4) [14] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com 7 | P a g e pppppppppppPhysical Sciences / P2 7 April 2021 Common Test Physical Sciences 7 April 2021 Common Test NSC-Guideline QUESTION 5 5.1.1 2,4-dimethylhexane (2) 5.1.2 1-chloro-2-methylpropane (2) 5.1.3 Esters  (1) 5.2.1 Compound that consists of carbon and hydrogen only.  (2) 5.2.2 Unsaturated. Contains a multiple bond/triple bond between carbon atoms in the carbon chain.  (2) 5.2.3 CnH2n-2 (1) 5.3 pentan-2-ol OR pentan-3-ol (2) 5.4.1 Compounds with same molecular formula but different functional groups.  (2) 5.4.2 H H H O  │ │ │ ║ H ― C ― C ― C ― C ― H │ │ │ H H H OR H H O  │ │ ║ H ― C ― C ― C ― H │ H H ― C ― H │ H OR H H O H │ │ ║ │ H ― C ― C ― C ― C ― H │ │ │ H H H  Whole structure correct: 2/2  Only functional group correct 1/2  More than one functional group 0/2 (2) 5.4.3 POSITIVE MARKING FROM Q5.4.2 butan-2-one OR butanal OR 2 - Methylpropanal (2) [18] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com 8 | P a g e pppppppppppPhysical Sciences / P2 8 April 2021 Common Test Physical Sciences 8 April 2021 Common Test NSC-Guideline QUESTION 6 6.1 The temperature at which the vapour pressure of a substance equals the atmospheric pressure.  (2) 6.2 163 °C (2) 6.3  The acid (A) has TWO sites for hydrogen bonding while the alcohol (B) has only one site.  The carboxylic acid molecules require more energy to overcome the intermolecular forces Carboxylic acid will have a higher boiling point.  (4) 6.4 Butanoic acid.  (2) 6.5 LOWER. _- CH3(CH2)2CH2OH or butan-1-ol has a shorter carbon chain/smaller surface area than compound B. Strength of the intermolecular forces in CH3(CH2)2CH2OH or butan-1-ol is weaker than that in compound B.  Therefore lesser energy needed to separate the molecules.  OR Compound B has a longer carbon chain/ larger surface area than CH3(CH2)2CH2OH or butan-1-ol The intermolecular forces between molecules of compound B are therefore strongerand require more energy to separate the molecules.  (3) [13] QUESTION 7 7.1.1 Hydrolysis (1) 7.1.2 Dehydrohalogenation (1) 7.2.1 Esterification (1) 7.2.2 Addition/hydration (1) 7.3.1 Heat Concentrated strong base in ethanol (2) 7.3.2 Propene/Prop-1-ene (1) 7.4 Water/H2O (1) 7.5.1 The carbon to which the hydroxyl (OH) group is bonded to, is bonded to ONE other carbon atom.  (2) 7.5.2 sulphuric acid/H2SO4 (1) 7.5.3 H O │ ║ H ― C ― C ― O ― H │ H  Whole structure correct: 2/2  Only functional group correct 1/2  More than one functional group 0/2 (2) [13] Copyright Reserved Please Turn Over