JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 20

Answer: (d) a, b, d

Explanation:
1. A secondary radical has its free valency on a carbon attached to two other carbons.
2. In (a) the radical carbon is bonded to CH3 and C2H5, in (b) to CH2= and CH3, and in (d) to two methyl groups.
3. In (c) the vinyl carbon is attached to one carbon only, so it is primary.

Teacher's Note:
1. Count the carbon neighbours of the radical carbon.
2. Ignore the bond order when counting neighbours.

Answer: (b) \( \ce{(CH3)3C-} \) : neobutyl

Explanation:
1. The group \( \ce{(CH3)3C-} \) is called tertiary butyl (tert-butyl), not neobutyl.
2. Neopentyl, propargyl and allyl are correctly matched to their groups.

Teacher's Note:
1. Learn the common names of unsaturated and branched groups.
2. Neobutyl is not an accepted name.

Answer: (b) 5

Explanation:
1. The principal chain must contain the carboxylic acid carbon (highest priority group) and should include the double bond.
2. The chain COOH-C=C-CH2-CH3 has 5 carbons.
3. The CHO is a substituent (formyl) and the propyl group is a branch, so the principal chain has 5 carbons.

Teacher's Note:
1. The principal chain contains the highest-priority functional group.
2. Count carbon atoms in the chain that includes the double bond.

Answer: (b) 4-Amino-3-methoxy-2-butenamide

Explanation:
1. The amide group has the highest priority, so the chain is numbered from the CONH2 carbon as C1.
2. C2 is =CH, C3 carries OCH3 and C4 carries the amino group (CH2NH2).
3. The name is 4-amino-3-methoxybut-2-enamide.

Teacher's Note:
1. Amide is the principal group and takes the suffix amide.
2. Amino and methoxy are only prefixes.

Answer: (c) N-Methyl ethanamine

Explanation:
1. The longer alkyl chain (ethyl) is the parent, giving ethanamine.
2. The other group on nitrogen is a methyl, which is shown with the locant N.
3. The name is N-methylethanamine.

Teacher's Note:
1. In a secondary amine the larger alkyl group is the parent chain.
2. The smaller group gets the locant N.

Answer: (b) Cyclohexane carbonyl chloride

Explanation:
1. The -COCl group is attached to a ring, so its carbon is not in the ring chain and the suffix is carbonyl chloride.
2. The name is cyclohexanecarbonyl chloride.

Teacher's Note:
1. When the functional group carbon is outside the ring, use carbonyl chloride, carboxylic acid or carbaldehyde.
2. Use oyl only for chain carbons.

Answer: (b) 2-Oxo ethane nitrile

Explanation:
1. The nitrile has higher priority than the aldehyde, so the parent is the two-carbon nitrile ethanenitrile.
2. The CHO group is shown as the prefix oxo on C2.
3. The name is 2-oxoethanenitrile.

Teacher's Note:
1. Nitrile outranks aldehyde in the priority order.
2. The carbon of the nitrile group is counted as C1.

Answer: (d) 3-Hydroxymethyl pentane-1,2,5-triol

Explanation:
1. The principal chain should contain the maximum number of OH groups, and then be the longest.
2. The five-carbon chain \( \ce{HOCH2-CH(OH)-CH(-)-CH2-CH2OH} \) contains three OH groups.
3. Numbering gives OH at 1, 2 and 5, with a hydroxymethyl group on C3, so the name is 3-(hydroxymethyl)pentane-1,2,5-triol.

Teacher's Note:
1. Choose the chain with the most principal groups first.
2. Number for the lowest locants of the principal group.

Answer: (c) 2-Methyl cyclopent-2-en-1-ol

Explanation:
1. The alcohol is the principal group, so the carbon bearing OH is C1.
2. Numbering goes towards the double bond, so the double bond is at C2=C3.
3. The methyl group is on C2, so the name is 2-methylcyclopent-2-en-1-ol.

Teacher's Note:
1. In rings the carbon with the principal group is C1.
2. Number through the double bond next.

Answer: (d) Ethanoic methanoic anhydride

Explanation:
1. The compound is a mixed anhydride of ethanoic acid and methanoic acid.
2. Mixed anhydrides are named by giving the two acid names in alphabetical order followed by anhydride.
3. The name is ethanoic methanoic anhydride.

Teacher's Note:
1. Anhydride outranks other groups in the name.
2. Name both acids in alphabetical order for mixed anhydrides.

Answer: (b) 2,3-Dimethyl butane

Explanation:
1. The longest chain has four carbons, \( \ce{CH3-CH(CH3)-CH(CH3)-CH3} \).
2. Methyl groups are on C2 and C3, so the name is 2,3-dimethylbutane.
3. 2,4-Dimethylbutane is not possible because C4 is a terminal carbon.

Teacher's Note:
1. Find the longest chain first.
2. Give the lowest locants to the substituents.

Answer: (d) Pent-3-en-1-yne

Explanation:
1. The chain has five carbons. When double and triple bonds are both present the lowest locants go to the multiple bonds together, and the triple bond gets the lower number if there is a choice.
2. Numbering from the triple bond end gives {1, 3}, and from the other end {2, 4}.
3. The name is pent-3-en-1-yne.

Teacher's Note:
1. Ene comes before yne in the name.
2. Use the lowest set of locants for all the multiple bonds.

Answer: (c) 3,4-Diethyl hexane

Explanation:
1. The longest chain has six carbons running through the two central carbons.
2. Each central carbon carries an ethyl group, at C3 and C4.
3. The name is 3,4-diethylhexane.

Teacher's Note:
1. Look for the longest chain, which may run through branches.
2. Count the carbons in each substituent separately.

Answer: (c) 3-Ethyl-2-methylpentane

Explanation:
1. In (a) the substituents are not in alphabetical order (ethyl must come before methyl).
2. In (b) an ethyl group on C2 means a longer six-carbon chain exists, and the compound is really 3,4-dimethylhexane. In (d) the same argument gives 3,3-dimethylhexane.
3. Only 3-ethyl-2-methylpentane (c) is correctly constructed: a five-carbon chain with methyl on C2 and ethyl on C3.

Teacher's Note:
1. An ethyl group on C2 of a chain means a longer chain exists.
2. Substituents are cited alphabetically.

Answer: (b) 5-Ethyl-3-methyl cyclohex-1-ene

Explanation:
1. The double bond carbons are C1 and C2.
2. Numbering through C2 towards the methyl group gives substituent locants {3, 5}, while the other direction gives {4, 6}.
3. The lower set {3, 5} is chosen: 5-ethyl-3-methylcyclohex-1-ene.

Teacher's Note:
1. The ring double bond takes locants 1 and 2.
2. Then choose the direction giving the lowest locants to the substituents.

Answer: (d) \( C_nH_{2n+2}O \)

Explanation:
1. Alkanols are alkanes in which one H is replaced by OH.
2. Alkane \( C_nH_{2n+2} \) becomes \( C_nH_{2n+1}OH = C_nH_{2n+2}O \).
3. For example ethanol is \( C_2H_6O \).

Teacher's Note:
1. Add one oxygen to the alkane formula.
2. Check with methanol CH4O.

Answer: (d) 2,3-Dimethylpentanoyl chloride

Explanation:
1. The longest chain containing COCl has five carbons, with the COCl carbon as C1.
2. Methyl groups are on C2 and C3.
3. The name is 2,3-dimethylpentanoyl chloride.

Teacher's Note:
1. The acid chloride carbon is always C1.
2. Number from the functional group end.

Answer: (a) 1-Buten-3-yne

Explanation:
1. The multiple bonds get the locant set {1, 3} from either end.
2. When there is a choice, the double bond gets the lower locant.
3. The name is but-1-en-3-yne.

Teacher's Note:
1. Ene is named before yne.
2. On a tie, the double bond gets the lower number.

Answer: (c) \( \ce{Br-CH2-CH=CH2} \) : 1-Bromoprop-2-ene

Explanation:
1. In \( \ce{BrCH2-CH=CH2} \) the double bond must get the lowest locant, so numbering starts from the \( \ce{CH2=} \) end.
2. The correct name is 3-bromoprop-1-ene, not 1-bromoprop-2-ene.
3. The other three names are correct.

Teacher's Note:
1. Multiple bonds get the lowest locants before halogen substituents.
2. Check numbering direction.

Answer: (c) \( \ce{HOOC-CH(OH)-C(CH2CH3)=C(CH3)-C#CH} \)

Explanation:
1. The parent is hexenynoic acid: C1 is COOH, the double bond is at C3 (C3=C4) and the triple bond is at C5 (C5#C6).
2. Substituents: OH on C2, ethyl on C3 and methyl on C4.
3. Only option (c) has this arrangement.

Teacher's Note:
1. Build the structure from the name step by step.
2. Start with the main chain, then the multiple bonds, then the substituents.

Answer: (b) Ethyne

Explanation:
1. The carbon in \( \ce{CO2} \) has two sigma bonds and no lone pair, so it is sp hybridised.
2. Ethyne has sp carbons (triple bond).
3. Ethane and ethanol have sp³ carbons, and ethene has sp² carbons.

Teacher's Note:
1. CO2 is linear with sp carbon.
2. Ethyne is the only sp compound here.

Answer: (d) 18 sigma bonds and 2 pi bonds

Explanation:
1. Count sigma bonds: C-H bonds are 3 (CH3) + 1 (=CH) + 5 (ethyl) = 9; C-C sigma bonds are 4 (CH3-C, C=C, C-C(O), CH2-CH3); C-O bonds are 4 (C-OH, C=O, C-O, O-CH2) and 1 O-H bond.
2. Total sigma bonds \( = 9 + 4 + 4 + 1 = 18 \).
3. Pi bonds are one C=C and one C=O, so 2.

Teacher's Note:
1. Every double bond has one sigma and one pi.
2. Count C-H, C-C, C-O and O-H bonds separately.