Hydrocarbons Test

Result

Hydrocarbons Test - 59

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Weekly Quiz Competition

Question 1
1 / -0

$$CH_4+Cl_2\xrightarrow{hv} CH_3Cl+HCl$$
To obtain high yields of $$CH_3Cl,$$ the ratio of $$CH_4$$ to $$Cl_2$$ must be:

A
high

B
low

C
equal

D
can't be predicted

Solution

$$CH_4+Cl_2 \xrightarrow [] {hv} CH_3Cl+HCl$$
To obtain high yields of $$CH_3Cl$$, the ratio of $$CH_4$$ to $$Cl_2$$ must be high, otherwise if we take low ratio of $$CH_4$$ to $$Cl_2$$ i.e. excess $$Cl_2$$ has been used then there will be formation of more than mono substituted products i.e. $$CH_2Cl_2, CHCl_3, CCl_4$$
Question 2
1 / -0

Arrange the following alkanes in decreasing order of their heats of combustion.
(i) $$\underset{(Neo-pentane) \, }{CH_3-\overset{CH_3}{\overset{|}{\underset{CH_3}{\underset{|}{C}}}}-CH_3}$$

(ii) $$\underset{(Iso-pentane)}{CH_3-\underset{CH_3}{\underset{|}{CH}}-CH_2-CH_3}$$

(iii)$$\underset{(n-pentane)}{CH_3-CH_2-CH_2-CH_2-CH_3}$$

A
(i) > (ii) > (iii)

B
(iii) > (I) > (ii)

C
(iii) > (ii) > (i)

D
(i) > (iii) > (ii)

Solution

$$CH_3$$
$$H_3C-\underset {|}{\overset {|}{C}}-CH_3$$ $$CH_3-\underset {|}{CH}-CH_2-CH_3$$ $$CH_3CH_2CH_2CH_2-CH_3$$
$$CH_3$$ $$CH_3$$
$$Neopentane (i)$$ $$Iso-pentane (ii)$$ $$n-pentane (iii)$$

We know that alkanes that are more branched, have less heat of combustion than a less branch one. Thus here heat of combustion order will be:
$$n-pentane(iii) > isopentane (ii) > Neopentane (i)$$.

Because here $$Neopentane$$ is more branched than $$isopentane$$ which is again more branched than $$n-pentane$$.
Question 3
1 / -0

$$CH_3-{\overset{CH_3}{\overset{|}C}}=CH_2 \, + \, (CH_3)_2CHCH_3 \, \xrightarrow[273K]{HF} \, C_8H_{18} \, (A)$$
Unknown $$(A)$$ in the above reaction is:

A
2,2,3-trimethyl pentane

B
2,2,4 -trimethyl pentane

C
2,2 dimethyl hexane

D
n-octane

Solution

The product is $$2,2,4-trimethyl$$ $$pentane$$
$$\rightarrow$$ First alkane attacks on acid and takes proton.
$$\rightarrow$$ $$F^{\circleddash}$$ abstract proton from alkane and form carbanion.
Question 4
1 / -0

Halogenation is a substitution reaction, where halogen replaces one or more hydrogens of hydrocarbon.
Chlorine free radical make $$1^{\circ} ,2^{\circ} ,3^{\circ}$$ radicals with almost equal ease, whereas bromine free radicals have a clear preference for the formation of tertiary free radicals. So , bromine is less reactive, and more selective whereas chlorine is less selective and more effective.
The relative rate of abstraction of hydrogen by Br is
$$\underset{(1600)}{3^\circ} \, > \, \underset{(82)}{2^\circ} \, > \, \underset{(1)}{1^\circ}$$
The relative rate of abstraction of hydrogen by Cl is
$$\underset{(5)}{3^\circ} \, > \, \underset{(3.8)}{2^\circ} \, > \, \underset{(1)}{1^\circ}$$
B.Above product will obtained in better yield if X is

A
$$Cl_2$$

B
$$I_2$$

C
$$Br_2$$

D
Can't be predicted

Solution

For the reaction $$R-H+X_2\overset {h\nu}{\longrightarrow} R-X+HX$$ to be obtained in better yeild, $$X$$ must be Bromine, This is because $$R$$ is a $$3^o$$ radical, hence we know relative rate of Bromination is very high.
Question 5
1 / -0

On halogenation, an alkane gives only one monohalogenated product. The alkane may be:

A
2-methyl butane

B
2, 2-dimethyl propane

C
cyclopentane

D
both (b) and (c)

Solution

A contains $$4$$ different $$H$$ atoms that's why A gives $$4$$ different mono halogenated product. But B and C have all equivalent $$H-$$atoms giving only one chlorinated (monosubstituted) product.
Question 6
1 / -0

In the reaction,
$$C_6H_6+CO+HCl \xrightarrow[AlCl_3]{anhyd.}X+HCl$$
The compound $$X$$ is:

A
$$C_6H_5CH_3$$

B
$$C_6H_5CH_2Cl$$

C
$$C_6H_5CHO$$

D
$$C_6H_5COOH$$
Question 7
1 / -0

In the preparation of alkanes, a concentrated aqueous solution of sodium or potassium salts of saturated carboxylic acid are subjected to:

A
hydrolysis

B
oxidation

C
hydrogenation

D
electrolysis

Solution

Carboxylic acids are electrolysed to form alkyl radical which combines to form alkane.
$$R-COO^- \rightarrow R-\overset{\overset{O}{||}}{C}-\dot{O}\rightarrow \dot{R}+CO_2$$
$$\dot{R}+\dot{R} \rightarrow R_2$$
Question 8
1 / -0

How many ionizable hydrogen's are there in the following compound?

A
2

B
4

C
5

D
7

Solution

Ionizable hyd are those where carbocation is stable:
Question 9
1 / -0

The barrier for rotation about the indicated bond is only $$14\ kcal/mol$$. This barrier is much lower than the barrier for rotation about $$C=C$$ bond in ethylene. The smaller barrier in this compound is due to:

A
dipolar structure

B
lack of resonance

C
enolization

D
combination of $$a$$ and $$b$$
Question 10
1 / -0

The arrangement of $${ \left( { CH }_{ 3 } \right) }_{ 3 }C-,{ \left( { CH }_{ 3 } \right) }_{ 2 }CH-,{ CH }_{ 3 }{ CH }_{ 2 }-$$ when attached to benzene or unsaturated group in increasing order of activating effect is:

A
$${ \left( { CH }_{ 3 } \right) }_{ 3 }C-<{ \left( { CH }_{ 3 } \right) }_{ 2 }CH-<{ CH }_{ 3 }{ CH }_{ 2 }-$$

B
$${ CH }_{ 3 }{ CH }_{ 2 }-<{ \left( { CH }_{ 3 } \right) }_{ 2 }CH-<{ \left( { CH }_{ 3 } \right) }_{ 3 }C-$$

C
$${ \left( { CH }_{ 3 } \right) }_{ 2 }CH-<{ \left( { CH }_{ 3 } \right) }_{ 3 }C-<{ CH }_{ 3 }{ CH }_{ 2 }-$$

D
$${ \left( { CH }_{ 3 } \right) }_{ 3 }C-<{ CH }_{ 3 }{ CH }_{ 2 }-<{ \left( { CH }_{ 3 } \right) }_{ 2 }CH-$$

Solution

The order of increasing activation order is
$${ \left( { CH }_{ 3 } \right) }_{ 3 }C-\quad <\quad { \left( { CH }_{ 3 } \right) }_{ 2 }CH-\quad <\quad { CH }_{ 3 }{ CH }_{ 2 }-$$
The greater the number of alkyl groups attached to a positive carbon atom, hyper conjugation increases and thus stability the cation.
As $${ \left( { CH }_{ 3 } \right) }_{ 3 }C-$$ has more alkyl groups attached it is more stable.