JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 50

Answer: (d) 1

Explanation:
1. Only 2-methylbutan-2-ol is a tertiary alcohol with this formula.

Teacher's Note:
1. Write out every possible structure systematically and avoid counting any twice.

Answer: (a) molecular formula

Explanation:
1. By definition, isomers have the same molecular formula but different structures, and so different properties.

Teacher's Note:
1. Write out every possible structure systematically and avoid counting any twice.

Answer: (c) chain isomers

Explanation:
1. Both are C\(_7\)H\(_{16}\) with different carbon skeletons, so they are chain isomers.

Teacher's Note:
1. Write out every possible structure systematically and avoid counting any twice.

Answer: (c) functional isomers

Explanation:
1. Both are C\(_3\)H\(_7\)NO.
2. One is a secondary amide (N-ethylformamide) and the other is an amino aldehyde, so they have different functional groups.

Teacher's Note:
1. Write out every possible structure systematically and avoid counting any twice.

Answer: (c) two primary alcohols, one secondary alcohol and one tertiary alcohol (all of these)

Explanation:
1. C\(_4\)H\(_{10}\)O has two primary alcohols (butan-1-ol, 2-methylpropan-1-ol), one secondary alcohol (butan-2-ol) and one tertiary alcohol (2-methylpropan-2-ol), apart from ethers.

Teacher's Note:
1. Write out every possible structure systematically and avoid counting any twice.

Answer: (b) metamers

Explanation:
1. Both are esters of formula C\(_9\)H\(_{10}\)O\(_2\) with different alkyl and acyl groups on the two sides of the ester linkage, so they are metamers.

Teacher's Note:
1. Write out every possible structure systematically and avoid counting any twice.

Answer: (b) functional group and general formula

Explanation:
1. A homologous series shares the functional group and general formula and differs by CH\(_2\) per member.
2. Empirical formula can differ.

Teacher's Note:
1. Write out every possible structure systematically and avoid counting any twice.

Answer: (c) Methanol and propan-1-ol

Explanation:
1. Methanol and propan-1-ol are both C\(_n\)H\(_{2n+2}\)O.
2. The other pairs belong to different general formulae.

Teacher's Note:
1. Write out every possible structure systematically and avoid counting any twice.

Answer: (c) 6,6-Dibromoheptan-2-ol

Explanation:
1. The longest chain has 7 carbons and -OH takes the lowest locant, so it is at C2.
2. The two bromines are then at C6.

Teacher's Note:
1. The principal functional group decides the direction of numbering.

Answer: (c) 3-Aminohept-5-enoic acid

Explanation:
1. The chain is numbered from COOH (C1) to give C1 COOH, C3 NH\(_2\) and a double bond at C5-C6.
2. The name is 3-aminohept-5-enoic acid.

Teacher's Note:
1. Write out every possible structure systematically and avoid counting any twice.

Answer: (b) 2-(Bromomethyl)-3-oxohexanamide

Explanation:
1. The amide carbon is C1 and the longest chain including it has six carbons.
2. C3 carries the oxo group and C2 carries CH\(_2\)Br, giving 2-(bromomethyl)-3-oxohexanamide.

Teacher's Note:
1. Write out every possible structure systematically and avoid counting any twice.

Answer: (c) Prop-2-enenitrile

Explanation:
1. The nitrile carbon is C1, so the chain is three carbons long with a double bond at C2: prop-2-enenitrile (acrylonitrile).

Teacher's Note:
1. Write out every possible structure systematically and avoid counting any twice.

Answer: (c) Ethyl 2-bromopropanoate

Explanation:
1. The alkyl group on oxygen (ethyl) is named first, and the acid part is a propanoate chain with bromine on C2.

Teacher's Note:
1. Write out every possible structure systematically and avoid counting any twice.

Answer: (c) Benzenecarbonyl chloride (benzoyl chloride)

Explanation:
1. An acyl chloride attached to a ring is named as ring + carbonyl chloride; the retained name is benzoyl chloride.

Teacher's Note:
1. Write out every possible structure systematically and avoid counting any twice.

Answer: (b) carboxylic acids

Explanation:
1. Saturated acids and esters have C\(_n\)H\(_{2n}\)O\(_2\).
2. Diketones and dialdehydes are C\(_n\)H\(_{2n-2}\)O\(_2\) and diols are C\(_n\)H\(_{2n+2}\)O\(_2\).

Teacher's Note:
1. Write out every possible structure systematically and avoid counting any twice.

Answer: (d) sp\(^2\) and sp\(^3\)

Explanation:
1. The carbonyl carbon is sp\(^2\) and the methyl carbon is sp\(^3\).

Teacher's Note:
1. Write out every possible structure systematically and avoid counting any twice.

Answer: (c) planar

Explanation:
1. The two sp\(^2\) carbons and the four atoms attached to them lie in one plane.

Teacher's Note:
1. Write out every possible structure systematically and avoid counting any twice.

Answer: (a) 120\(^\circ\)

Explanation:
1. Each carbon is sp\(^2\) hybridised, giving trigonal planar geometry with bond angles near 120\(^\circ\).

Teacher's Note:
1. Write out every possible structure systematically and avoid counting any twice.

Answer: (a) 6 sigma and 2 pi

Explanation:
1. There are 4 C-H bonds and 2 C-C sigma bonds (6 sigma), and one pi bond in each C=C (2 pi).

Teacher's Note:
1. Write out every possible structure systematically and avoid counting any twice.

Answer: (d) H\(_2\)C=CH-CN

Explanation:
1. In H\(_2\)C=CH-C\(\equiv\)N the carbons are sp\(^2\), sp\(^2\), sp and the nitrogen is sp.
2. The others do not follow this order.

Teacher's Note:
1. Write out every possible structure systematically and avoid counting any twice.

Answer: (d) aspirin

Explanation:
1. Aspirin has a free COOH group.
2. Picric acid is a phenol, barbituric acid is a cyclic amide and ascorbic acid has a lactone and enol groups.

Teacher's Note:
1. Write out every possible structure systematically and avoid counting any twice.

Answer: (c) B and C

Explanation:
1. Substituents are cited alphabetically, so chloro comes before methyl in 1-chloro-4-methylbenzene (C), and toluene is an accepted parent name (B).

Teacher's Note:
1. Write out every possible structure systematically and avoid counting any twice.

Answer: (d) A, B and C are incorrect statements

Explanation:
1. Aniline (C\(_6\)H\(_5\)NH\(_2\)) is a homocyclic aromatic compound, not heterocyclic (A is wrong).
2. It has 14 sigma bonds (B is wrong), and DU = 4 (C is wrong).
3. It does contain an amine group (D is right).

Teacher's Note:
1. Count every C-H, C-C and N-H bond for the sigma total.

Answer: (d) B and C only

Explanation:
1. C\(_3\)H\(_8\) has one structure only (A false).
2. C\(_5\)H\(_{12}\) has 3 and C\(_6\)H\(_{14}\) has 5 (B, C true).
3. C\(_6\)H\(_4\)BrCl has 3 benzene isomers: o-, m- and p- (D false).

Teacher's Note:
1. Write out every possible structure systematically and avoid counting any twice.