Home chapter1 Alkanes:Que

chapter1 Alkanes:Que

Ruhul Amin Alig

Question:

$_{C2} H_{5} C l + N a D r y e t h e r - ----- \to X + N a C l$ Number of possible structural and stereo isomers for mono chloro derivatives of 'X' are

OPTIONS

Solution :

$C_{4} H_{10}$ has 2 structural and one chiral centre so totally 3 isomers.

Answer is B

Question.

Solution :

Let us inspect option A

Above products on oxidation yield

O O O O

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C H_{3} - C - H O, C H_{3} - C - C - O H, C H_{3} - C - C H_{3} (u n a f f e c t e d)

Let us also investigate (b)

C (C H_{3})_{2}

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C H_{3} - C - C H = C H - C H_{3}

Above products on oxidation yield,

O O O O

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C H_{3} - C - C H_{3} (u n a f f e c t e d), C H_{3} - C - C - O H a n d C H_{3} - C - O H

So, both options (a) and (b) are correct.

Question:

Select one or more options

CO₂

HCHO

H₂O

CH₃OH

Solution :

This is one of the examples of combustion of alkanes.

Alkanes readily burn in excess of air or oxyge to give $C O_{2}$ and $H_{2} O$

i. $C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$

ii. $C_{3} H_{8} + 5 O_{2} \to 3 C O_{2} + 4 H_{2} O$

iii. $C_{n} H_{2 n + 2} + (\frac{3 n + 1}{2}) O_{2} △ \to n C O_{2} + (n + 1) H_{2} O + H e a t (Answer is option A and C)$

Questin:What is the chief product obtained when n-butane is treated with bromine in the presence of light at $130^{\circ} C$

OPTIONS

Solution :

Remember Bromination of alkane.

$C H_{3} - C H_{2} - C H_{2} - C H_{3} + B r_{2} h ν \to$

2-Bromobutane is the main product because $2^{\circ}$ carbonium ion is more stable than $1^{\circ}$ .

answer is A

Question:

$H_{3} C \equiv C H N a N H_{2} - ---- \to X C H_{2} C H_{2} B r - ------ \to Y .$ Identify the product 'Y' is

OPTIONS

Pent - 2 - ene

Pent - 2 - yne

Solution :

X is $C H_{3} - C \equiv C^{-} N a^{+}$

Y is $C H_{3} - C \equiv C - C H_{2} - C H_{3}$

The first step removes the terminal hydrogen from the given substrate (reactant on the left). Such a removable hydrogen is called an acidic hydrogen. In the next step, the nucleophile - which is rich in electrons - attacks the $C - B r$ bond of ethyl bromide. Because of the difference in electronegativity between carbon and bromine, this $C - B r$ bond is polar - meaning, the bonding electrons are closer to the more electronegative bromine. This gives rise to the possibility of the carbon of the $C - B r$ bond having a slight deficit of electrons; let's just say that this carbon misses electrons and that is where the electron-rich terminal carbon of $C H_{3} - C \equiv C^{-}$ comes in. Hence, the reaction.