Here is an post-mortem analysis / "how to" for the MT. The questions are split by the sections. At the start of each section are a few suggestions of what to look for or how to tackle the question type.

Qu1: A
Question about geometry goes back to hybridisation.  i is an alkene so the central C is sp² , i.e. trigonal planar, with bond angles of 120^o, ii  is an alkane so the central C is sp³ , i.e. trigonal planar, with bond angles of 109^o, and iii  is a cyclobutane so the central C is sp³ , but the ring forces the angle down to almost 90^o, therefore giving : i > ii > iii

Qu2:  AB
Since all are CH bonds we should look at the hybridisation of the C atom.... the greater the s character of the C hybrid used to form the sigma bond, the shorter and the stronger that bond will be. Here we have an alkane (sp³), an alkene (sp2) and an alkyne (sp)... therefore iii > ii > i

Qu3: A
Based on the rules for ranking resonance structures, look at the octets and then charge separation.  i has complete octets at C, N and O and the negative charge on the more electronegative O atom. ii also has complete octets at C, N and O but now the negative charge on the less electronegative C atom. In iii there N has violated the octet rule by having 10 electrons around it, which it can't do. Therefore, in order of importance we have i > ii > iii

Qu4: C
Calculate the formal charges for the N atom = group number - number of bonds - lone pairs electrons : i = neutral based on 5 - 3 - 2, ii = +1 (draw it !) based on 5 - 4  and iii = -1 based on 5 - 2 - 4. Therefore ii > i > iii

Qu5: E
The most stable form (i.e. low energy conformation) of cyclohexane is the chair ii. i shows a boat conformation and the iii the half chair which is quite unstable due to the planar nature for 5 of the 6 C atoms. this planarity introduces eclipsing interactions and torsional strain and is actually the high energy point on the conversion of cyclohexane from chair to chair.  So we have iii > i > ii

Qu6: D
More branched isomers are more stable due to increased intramolecular Van der Waals contacts and maximising the number of stronger primary CH bonds. the more stable the isomer the less exothermic the heat of combustion. Therefore ii > iii > i

Qu7: B
Just have to draw them out, no quick way.  Constitutional isomers means different numbers and types of bonds giving different functional groups and branching patterns.
C₃H₆O could be propanal, propanone, cyclopropanol, 1-propen-1-ol, 1-propen-2-ol, 3-propen-1-ol, ethenyl methyl ether, propylene oxide, oxacyclobutane (9 isomers).
C3H6 could be propene or cyclopropane (2 isomers).
C₆H₁₄ can be a whole bunch of things.... hexane, 2-methylpentane, 3-methylpentane, 2,2-dimethylbutane, 2,3-dimethylbutane (5 isomers)
So i > iii > ii

Qu8: A
The greater the s character in the hybrid then the lower the energy of that orbital. sp³ is 25%, sp² is 33% and sp is 50% so therefore in terms of the energies, i > ii > iii

Qu9: AB

Qu10: C

Qu11: B

Qu12: AC

Qu13: E

Qu14: A

Qu15: CD

Qu16: AE
Water needs to go in at the bottom and out at the top in order to use gravity to ensure that the condenser is full of water to facilitate cooling.

Qu17: AD
See above.

Qu18: D

Qu19: A
C3 is attached to 2 x O = -2, 1 x N = -1 and 1 x C = 0 so the sum = -3, therefore C3 = +3

Qu20: E
C17 is attached to 3 x H = +3 and 1 x O = -1 so the sum = +2, therefore C17 = -2

Qu21: D
Since all are CC bonds we need to consider the hybridisation of the atoms involved and the type of bond. A double bond will be shorter than a single bond due to the increased bonding interaction. C1-C3 is a single bond sp³ to sp², C5-C6 is a single bond sp³ to sp³, C6-C7 is a single bond sp³ to sp², C7-C8 is a double bond sp² to sp² C11-C12 is sp² to sp² (benzene so between C-C and C=C in character). Therefore C7-C8 will be the shortest.

Qu22: C
The proximity of the N and C=O system gives the amide group.

Qu23: C
The C-O-C system (no C=O there) implies an ether.

Qu24: C
Count the rings and the p bonds to get 2 rings plus 5 p bonds = 7

Qu25: C
O2 is sp², because it is involved in the double bond (1 attached group, 2 lone pairs). N4 is sp², because the lone pair is involved in resonance with the C=O double bond (it is an amide). C5 is sp³ (4 attached groups, 2 C and the 2 undrawn H to assume). N9 is sp², because the lone pair is involved in resonance with the C=C double bond and C11 is sp² (3 attached groups all C).  Most likely you overlooked the effect of resonance and the requirement to put the lone pair in a p orbital for this.

Qu26: C
The diagram shows a Newman projection of cyclohexane in the chair conformation. Note that the two -CH2- groups drawn are one up and one down, and that the sides are in a staggered conformation.

Qu27: D
The methyl group is on a bond in the plane of the ring (equatorial) while the chlorine is on a bond perpendicular to that plane (axial).

Qu28: B
Isomers that have different functional groups or branching are constitutional isomers.

Qu29: CD
To be conformational isomers of methylcyclohexane they need that name. A is a cyclohexene system, B is a cyclopentane, C is a boat conformation of methylcyclohexane, D is a chair conformation of methylcyclohexane and E is the half chair conformation of just cyclohexane.

Qu30: B
A  is ring strain, B is torsional strain, C is angle strain, D and E are both steric interactions.

Qu31: D
First draw it out.  To be the most stable conformation it needs to be staggered with the large t-butyl group away from the methyl group. B and E are both eclipsed. A and C are both staggered but with the t-butyl group gauche to methyl groups so that leaves D where the t-butyl is as far away from the methyl group as possible in an anti position.

Qu32: C
An alcohol, -OH has priority low number gives C3, so it's an "-3-ol" with an alkene that is E (i.e. groups opposite across the C=C). 7 carbon chain so a "hepten-3-ol"

Qu33: B
A ketone, C=O that has priority so it's an "-one". With the C=O at C1, the lowest number for the methyl groups is C3, two of them so 3,3-dimethyl.

Qu34: B
An ether. Two different groups attached. The benzyl on the left and a phenyl on the right.

Qu35: B
Complex substituents in an alkane... locate longest chain, here C10, identify substituents, number based on first point of difference (i.e. 4,4- groups rather than just a 4- group). The complex substituent has a C2 chain (so ethyl) with a methyl group at C1 so 1-methylethyl located at C7 on the main chain.... Remember to alphabetise....

Qu36: E
Use the descriptors and look at the -OH group.... A is Z but has no R or S and the -OH is at C3. B and C are E but have no R or S nor the -OH. D is (2Z,5R) and E is (2Z,5S).

Qu37: D
An ester based on benzene narrows it to D or E but E is a methyl ester.

Qu38: A
Use the descriptors.... first the N,N- means we have two methyl groups both on the N atom, so C must be wrong. Its also an aldehyde ("-al") so B and D are out. Down to A and E.  Priority order for groups at the chirality center is N > C=O > ethyl > H. A is S, E is R.

Qu39: C
Did you look at bicyclics ? Bicyclic means two fused rings. [2.2.1] means that there are 2C, 2C and 1C in the links between the shared C atoms. This gets rid of E. We number from one of the shared C atoms via the longest link first so as to give the alkene the lowest number.

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