Organic Chemistry – Some Basic Principles and Techniques Chapter-Wise Test 11

Correct answer Carries: 4.

Wrong Answer Carries: -1.

In the Dumas method, 0.28 g of a compound gave 60 mL of \( \ce{N2} \) at STP. If the compound contains only C, H, and N, what is the percentage of nitrogen?

Mass of N = \( \frac{28}{22400} \times 60 = 0.075 \) g. Percentage = \( \frac{0.075}{0.28} \times 100 = 26.79\% \approx 27\% \).

25%
26%
28%
27%
4

Which method is used to separate a mixture of two liquids with different boiling points?

Distillation separates liquids based on differences in their boiling points, vaporizing the lower boiling point liquid first.

Sublimation
Crystallization
Distillation
Chromatography
3

Which type of bond cleavage produces a carbanion?

Heterolytic cleavage, where the bond breaks unequally, can produce a carbanion (e.g., \( \ce{CH3^-} \)) if the carbon gains both electrons.

Homolytic
Heterolytic
Radical
Covalent
2

What is the IUPAC name of \( \ce{CH3CH2CH(OH)CH3} \)?

The compound has a 4-carbon chain (butane) with an -OH on the second carbon, named butan-2-ol.

Butan-2-ol
Butan-1-ol
Propan-2-ol
2-Methylpropan-1-ol
1

Which of the following carbocations is least stable?

Stability decreases with less substitution: primary < secondary < tertiary. \( \ce{CH3CH2+} \) (primary) is the least stable.

\( \ce{(CH3)3C+} \)
\( \ce{CH3CH+CH3} \)
\( \ce{CH3CH2+} \)
\( \ce{(CH3)2CHCH2+} \)
3

What is the shape of the carbon atom in \( \ce{CH3CH2C#CH} \) attached to the methyl group?

In \( \ce{CH3CH2C#CH} \), the carbon in \( \ce{CH3} \) has four single bonds (three H, one C), indicating \( sp^3 \) hybridization and a tetrahedral shape.

Linear
Trigonal planar
Tetrahedral
Bent
3

What is the shape of the carbon atom in \( \ce{CH3C#N} \) that is triple-bonded to nitrogen?

The carbon in \( \ce{CH3C#N} \) triple-bonded to nitrogen has two sigma bonds (C-C and C≡N) and is \( sp \) hybridized, resulting in a linear shape.

Trigonal planar
Tetrahedral
Linear
Bent
3

How many \( \sigma \) bonds are present in \( \ce{CH2=CH-CH3} \)?

In \( \ce{CH2=CH-CH3} \) (propene), there are 8 single bonds (C-H and C-C) and 1 double bond (C=C). The double bond consists of 1 \( \sigma \) and 1 \( \pi \), so total \( \sigma \) bonds = 8 (single) + 1 (from double) = 9.

9
8
10
7
1

Which species is most likely to act as a nucleophile in the conversion of \( \ce{CH3CH2Br} \) to \( \ce{CH3CH2CN} \)?

\( \ce{CN^-} \) (cyanide ion) is a strong nucleophile that displaces \( \ce{Br^-} \) in a substitution reaction to form \( \ce{CH3CH2CN} \) (propanenitrile).

\( \ce{Br^-} \)
\( \ce{H2O} \)
\( \ce{CH3CH2+} \)
\( \ce{CN^-} \)
3

Which carbocation is the most stable among the following?

Tertiary carbocations are more stable due to hyperconjugation and inductive effects. \( \ce{(CH3)3C+} \) (tertiary) is more stable than primary or secondary ones.

\( \ce{CH3CH2CH2+} \)
\( \ce{CH3CH+CH3} \)
\( \ce{(CH3)3C+} \)
\( \ce{CH3+} \)
3

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