The C – H bond length is minimum in the bond formed by

sp- s overlapping (as in alkynes)
sp²– s overlapping (as in alkenes)
sp³– s overlapping (as in alkanes)
None of these

a
b
c
d

Solution

Methyl alcohol
Acetaldehyde
C₂H₅OH
C₂H₄

Solution

I
II
III
all are equal electronegative

Solution

As the number of bonds between carbon atoms increases, electronegativity of that carbon also increases due to increasing active power of electrons.Also sp hybrid is more e–negetive than sp² which is more electronegative than sp³(s character decreases)Hence, option (a) is correct.

(CH₃)₂C = CHCH₃+ NOBr ⟶ product. The structure of the product is

(CH₃)₂ C(NO) – CH(Br)CH₃
(CH₃)₂ C(Br) – CH(NO)CH₃
(CH₃)₂ CH – C(NO)(Br)CH₃

Solution

The reaction follows Markownikoff rule, namely the bromide ion adds on to the carbon having the least number of hydrogen and the more positive part namely the – NO group adds to the other carbon of the double bond.

Identify the reagent from the following list which can easily distinguish between 1-butyne and 2-butyne

bromine, CCl₄
H₂, Lindlar catalyst
dilute H₂SO₄, HgSO₄
ammonical Cu₂Cl₂ solution

Solution

In 1-butyne terminal hydrogen is acidic where as in 2-butyne there is no terminal hydrogen.Thus 2-butyne will not react with ammonical Cu₂Cl₂. While 1-butyne,being terminal alkyne, will give red ppt. with ammonical cuprous chloride.

When acetylene passed through dil. H₂SO₄ in presence of HgSO4, the compound formed is

ether
acetaldehyde
acetic acid
ketone

Solution

Primary amine
An amide
Phenyl isocyanate
Achain lengthened hydrocarbon

Solution

Ammonical silver nitrate forms a white precipitate easily with

CH₃C ≡ CH
CH₃C = CCH₃
CH₃CH = CH₂
CH₂ = CH₂

Solution

Terminal alkyenes give a white precipitate easily on reaction with ammonical silver nitrate solution.

In the reaction CH₃ = C^– – Na⁺ + (CH₃)₂CH –
Cl?
The product formed is :

propene
propyne
propyne and propene
4-methylpentyne-2

Solution

Attempted

0

Correct

0

Wrong

0

Attempted

Wrong

Correct