Organic Chemistry: Some Basic Principal & Technique Chapter-Wise Test 1

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Which free radical formed from \( \ce{CH3CH2CH2CH2CH3} \) via homolytic cleavage is most stable?

Cleavage at \( \ce{CH2-CH2} \) forms \( \ce{CH3CH2CH2CH2CH2.} \) (primary) or at \( \ce{CH2-CH} \) forms \( \ce{CH3CH2CH.CH2CH3} \) (secondary), which is more stable due to hyperconjugation.

\( \ce{CH3CH2CH2CH2.} \)
\( \ce{CH3CH2.} \)
\( \ce{CH3CH2CH.CH2CH3} \)
\( \ce{CH3CH2CH2CH2CH2.} \)
3

What is the hybridization state of the carbon atom in the carbonyl group of \( \ce{CH3CHO} \)?

The carbonyl carbon in \( \ce{CH3CHO} \) (acetaldehyde) is bonded to three atoms: one oxygen (double bond), one carbon, and one hydrogen. This involves \( sp^2 \) hybridization, forming a trigonal planar shape.

\( sp^2 \)
\( sp^3 \)
\( sp \)
\( sp^3d \)
1

In Lassaigne’s test, the presence of nitrogen gives a Prussian blue color due to the formation of which compound?

In Lassaigne’s test, nitrogen forms \( \ce{Fe4[Fe(CN)6]3} \) (ferric ferrocyanide) with \( \ce{Fe^{3+}} \), producing the Prussian blue color.

\( \ce{Fe4[Fe(CN)6]3} \)
\( \ce{Na4[Fe(CN)6]} \)
\( \ce{Fe2[Fe(CN)6]4} \)
\( \ce{Fe(CN)6} \)
1

What is the hybridization of the carbon atom in \( \ce{CH3CH2C#N} \) attached to the ethyl group?

In \( \ce{CH3CH2C#N} \), the carbon in \( \ce{CH2} \) has four single bonds (two H, one C to \( \ce{CH3} \), one C to \( \ce{C#N} \)), indicating \( sp^3 \) hybridization.

\( sp \)
\( sp^3 \)
\( sp^2 \)
\( sp^3d \)
2

Which intermediate is most likely to form during homolytic cleavage of \( \ce{CH3CH2CH2CH2Br} \) under UV light?

Homolytic cleavage under UV light splits \( \ce{C-Br} \) equally, forming \( \ce{CH3CH2CH2CH2.} \) (butyl radical) and \( \ce{Br.} \), typical in radical reactions.

\( \ce{CH3CH2CH2CH2+} \)
\( \ce{CH3CH2CH2CH2.} \)
\( \ce{CH3CH2CH2CH2^-} \)
\( \ce{Br+} \)
2

Which functional group is present in \( \ce{CH3CH2COOH} \)?

\( \ce{CH3CH2COOH} \) (propanoic acid) contains the \( \ce{-COOH} \) group, characteristic of carboxylic acids.

Aldehyde
Ketone
Ester
Carboxylic acid
4

What is the total number of \( \pi \) bonds in \( \ce{CH3C#CCH3} \)?

In \( \ce{CH3C#CCH3} \) (but-2-yne), the triple bond (\( \ce{C#C} \)) consists of 2 \( \pi \) bonds.

2
1
3
4
1

An organic compound with two \( \pi \) bonds has a total of 7 sigma bonds. What is its molecular formula?

Two \( \pi \) bonds suggest a diene or alkyne. \( \ce{CH2=CH-CH=CH2} \) (buta-1,3-diene) has 2 \( \pi \) bonds (two C=C) and 7 \( \sigma \) bonds (5 C-H, 2 C-C), fitting \( \ce{C4H6} \).

\( \ce{C3H4} \)
\( \ce{C4H6} \)
\( \ce{C5H8} \)
\( \ce{C4H8} \)
2

Which type of isomerism is shown by \( \ce{CH3CH2CHO} \) and \( \ce{CH3COCH3} \)?

\( \ce{CH3CH2CHO} \) (propanal) and \( \ce{CH3COCH3} \) (acetone) have the same molecular formula (\( \ce{C3H6O} \)) but different functional groups (aldehyde vs. ketone), indicating functional isomerism.

Chain isomerism
Functional isomerism
Position isomerism
Geometrical isomerism
2

Which method is used to estimate halogens in an organic compound by heating with fuming nitric acid?

The Carius method involves heating the compound with fuming \( \ce{HNO3} \) to convert halogens to silver halides.

Dumas method
Kjeldahl’s method
Carius method
Lassaigne’s test
3

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