Q. No.(CH3)2CHCH(Cl)CH3 can be classified as:
1. Secondary benzyl halide
2. Secondary alkyl halide
3. Primary alkyl halide
4. Secondary allyl halide
(CH3)3CCH2CH(Br)C6H5 is:
1. Secondary benzyl halide
2. Secondary alkyl halide
3. Primary alkyl halide
4. Secondary allyl halide
CH3CH2CH(CH3)CH(C2H5)Cl is:
1. Secondary benzyl halide
2. Secondary alkyl halide
3. Primary alkyl halide
4. Secondary allyl halide
CH3CH2C(CH3)2CH2I is:
1. Secondary benzyl halide
2. Secondary alkyl halide
3. Primary alkyl halide
4. Secondary allyl halide
The correct statement among the following about C5H11Br is:
| 1. | It has total 8 isomers, 4 are primary alkyl bromide isomers and 2 are tertiary alkyl bromide isomers. |
| 2. | It has total 7 isomers, 4 are primary alkyl bromide isomers and 2 are tertiary alkyl bromide isomers. |
| 3. | It has total 8 isomers, 4 are primary alkyl bromide isomers and 3 are secondary alkyl bromide isomers. |
| 4. | It has total 7 isomers, 3 are secondary alkyl bromide isomers and 1 is a tertiary alkyl bromide isomer. |
The number of monochloro structural isomers formed on free radical monochlorination of (CH3)2CHCH2CH3 is:
1. 5
2. 3
3. 4
4. 6
Which of the carbon atoms present in the molecule given below are asymmetric?
1. 1, 2, 3, 4
2. 2, 3
3. 1, 4
4. 1, 2, 3
Compound(s) among the following with an asterisk (*) asymmetric carbon is/are:
1. (a), (b), (c), (d)
2. (a), (b), (c)
3. (b), (c), (d)
4. (a), (c), (d)
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