NEET Test Series from KOTA - 10 Papers In MS WORD
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Hydrocarbons
230281
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_3 \stackrel{400-600^{\circ} \mathrm{C}}{\longrightarrow} \mathrm{X}+\mathrm{Y}$ Xand $Y$ are
1 Hydrogen
2 Methane
3 Hydrogen
4 Ethylene
Explanation:
: $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_3 \stackrel{400-600^{\circ} \mathrm{C}}{\longrightarrow} \underset{(\mathrm{X})}{\mathrm{CH}_4}+\mathrm{CH}_2=\mathrm{CY}_{(\mathrm{Y})} \mathrm{CH}_2$ In this reaction $\mathrm{X}$ is methane and $\mathrm{Y}$ is ethylene. Hence, the option $\mathrm{b}$ and $\mathrm{d}$ are correct.
UPTU/UPSEE-2008
Hydrocarbons
230282
In cyclopropane, cyclobutane and cyclohexane, the common group is
1
2
3
4
Explanation:
: The structure of the cyclobutane, cyclobutane and cyclohexane are - In this structure we can observe that common structure present in all hydrocarbon is
UPTU/UPSEE-2008
Hydrocarbons
230283
How many monobriminated product (s) (including stereoisomers) would form in the free radical bromination of n-butane?
1 2
2 1
3 3
4 4
Explanation:
: In n-butane monobrominated of product will 3 (including stereoisomer's).
WB-JEE-30.04.2022
Hydrocarbons
230286
The $4^{\text {th }}$ higher homologue of ethane is
1 Butane
2 Pentane
3 Hexane
4 Heptane
Explanation:
: We know that the molecular formula of ethane is $\mathrm{C}_2 \mathrm{H}_6$. The $4^{\text {th }}$ higher homologue is - $4 \times \mathrm{CH}_2=\mathrm{C}_4 \mathrm{H}_8$ $\therefore$ The fourth $4^{\text {th }}$ homologue of ethane is $\mathrm{C}_2 \mathrm{H}_6+\mathrm{C}_4 \mathrm{H}_8=\mathrm{C}_6 \mathrm{H}_{14}$ (Hexane). Thus, hexane is the fourth higher homologue of ethane.
230281
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_3 \stackrel{400-600^{\circ} \mathrm{C}}{\longrightarrow} \mathrm{X}+\mathrm{Y}$ Xand $Y$ are
1 Hydrogen
2 Methane
3 Hydrogen
4 Ethylene
Explanation:
: $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_3 \stackrel{400-600^{\circ} \mathrm{C}}{\longrightarrow} \underset{(\mathrm{X})}{\mathrm{CH}_4}+\mathrm{CH}_2=\mathrm{CY}_{(\mathrm{Y})} \mathrm{CH}_2$ In this reaction $\mathrm{X}$ is methane and $\mathrm{Y}$ is ethylene. Hence, the option $\mathrm{b}$ and $\mathrm{d}$ are correct.
UPTU/UPSEE-2008
Hydrocarbons
230282
In cyclopropane, cyclobutane and cyclohexane, the common group is
1
2
3
4
Explanation:
: The structure of the cyclobutane, cyclobutane and cyclohexane are - In this structure we can observe that common structure present in all hydrocarbon is
UPTU/UPSEE-2008
Hydrocarbons
230283
How many monobriminated product (s) (including stereoisomers) would form in the free radical bromination of n-butane?
1 2
2 1
3 3
4 4
Explanation:
: In n-butane monobrominated of product will 3 (including stereoisomer's).
WB-JEE-30.04.2022
Hydrocarbons
230286
The $4^{\text {th }}$ higher homologue of ethane is
1 Butane
2 Pentane
3 Hexane
4 Heptane
Explanation:
: We know that the molecular formula of ethane is $\mathrm{C}_2 \mathrm{H}_6$. The $4^{\text {th }}$ higher homologue is - $4 \times \mathrm{CH}_2=\mathrm{C}_4 \mathrm{H}_8$ $\therefore$ The fourth $4^{\text {th }}$ homologue of ethane is $\mathrm{C}_2 \mathrm{H}_6+\mathrm{C}_4 \mathrm{H}_8=\mathrm{C}_6 \mathrm{H}_{14}$ (Hexane). Thus, hexane is the fourth higher homologue of ethane.
230281
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_3 \stackrel{400-600^{\circ} \mathrm{C}}{\longrightarrow} \mathrm{X}+\mathrm{Y}$ Xand $Y$ are
1 Hydrogen
2 Methane
3 Hydrogen
4 Ethylene
Explanation:
: $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_3 \stackrel{400-600^{\circ} \mathrm{C}}{\longrightarrow} \underset{(\mathrm{X})}{\mathrm{CH}_4}+\mathrm{CH}_2=\mathrm{CY}_{(\mathrm{Y})} \mathrm{CH}_2$ In this reaction $\mathrm{X}$ is methane and $\mathrm{Y}$ is ethylene. Hence, the option $\mathrm{b}$ and $\mathrm{d}$ are correct.
UPTU/UPSEE-2008
Hydrocarbons
230282
In cyclopropane, cyclobutane and cyclohexane, the common group is
1
2
3
4
Explanation:
: The structure of the cyclobutane, cyclobutane and cyclohexane are - In this structure we can observe that common structure present in all hydrocarbon is
UPTU/UPSEE-2008
Hydrocarbons
230283
How many monobriminated product (s) (including stereoisomers) would form in the free radical bromination of n-butane?
1 2
2 1
3 3
4 4
Explanation:
: In n-butane monobrominated of product will 3 (including stereoisomer's).
WB-JEE-30.04.2022
Hydrocarbons
230286
The $4^{\text {th }}$ higher homologue of ethane is
1 Butane
2 Pentane
3 Hexane
4 Heptane
Explanation:
: We know that the molecular formula of ethane is $\mathrm{C}_2 \mathrm{H}_6$. The $4^{\text {th }}$ higher homologue is - $4 \times \mathrm{CH}_2=\mathrm{C}_4 \mathrm{H}_8$ $\therefore$ The fourth $4^{\text {th }}$ homologue of ethane is $\mathrm{C}_2 \mathrm{H}_6+\mathrm{C}_4 \mathrm{H}_8=\mathrm{C}_6 \mathrm{H}_{14}$ (Hexane). Thus, hexane is the fourth higher homologue of ethane.
230281
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_3 \stackrel{400-600^{\circ} \mathrm{C}}{\longrightarrow} \mathrm{X}+\mathrm{Y}$ Xand $Y$ are
1 Hydrogen
2 Methane
3 Hydrogen
4 Ethylene
Explanation:
: $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_3 \stackrel{400-600^{\circ} \mathrm{C}}{\longrightarrow} \underset{(\mathrm{X})}{\mathrm{CH}_4}+\mathrm{CH}_2=\mathrm{CY}_{(\mathrm{Y})} \mathrm{CH}_2$ In this reaction $\mathrm{X}$ is methane and $\mathrm{Y}$ is ethylene. Hence, the option $\mathrm{b}$ and $\mathrm{d}$ are correct.
UPTU/UPSEE-2008
Hydrocarbons
230282
In cyclopropane, cyclobutane and cyclohexane, the common group is
1
2
3
4
Explanation:
: The structure of the cyclobutane, cyclobutane and cyclohexane are - In this structure we can observe that common structure present in all hydrocarbon is
UPTU/UPSEE-2008
Hydrocarbons
230283
How many monobriminated product (s) (including stereoisomers) would form in the free radical bromination of n-butane?
1 2
2 1
3 3
4 4
Explanation:
: In n-butane monobrominated of product will 3 (including stereoisomer's).
WB-JEE-30.04.2022
Hydrocarbons
230286
The $4^{\text {th }}$ higher homologue of ethane is
1 Butane
2 Pentane
3 Hexane
4 Heptane
Explanation:
: We know that the molecular formula of ethane is $\mathrm{C}_2 \mathrm{H}_6$. The $4^{\text {th }}$ higher homologue is - $4 \times \mathrm{CH}_2=\mathrm{C}_4 \mathrm{H}_8$ $\therefore$ The fourth $4^{\text {th }}$ homologue of ethane is $\mathrm{C}_2 \mathrm{H}_6+\mathrm{C}_4 \mathrm{H}_8=\mathrm{C}_6 \mathrm{H}_{14}$ (Hexane). Thus, hexane is the fourth higher homologue of ethane.
