Reaction of HBr with propene in the presence of peroxide gives n-propyl bromide.
This addition reaction is an example of anti-Markovnikov’s addition reaction.
Reaction of HBr with propene in the presence of peroxide gives n-propyl bromide.
This addition reaction is an example of anti-Markovnikov’s addition reaction.
With ethene in presence of anhydrous AlCl3, benzene gives ethyl benzene (PhEt).
–NH 2 group is electron donating hence increases electron density on ring.
Benzene is also electron rich due to delocalisation of electrons. –NO 2 group is electron withdrawing hence, decreases electron density on ring.
Thus, correct order for electrophilic substitution is I II III.
Due to restricted rotation about double bond, 2-butene shows geometrical isomerism.
Propene adds to diborane (B 2 H 6 ) giving an addition product.
The addition compound on oxidation gives 1-propanol.
Here addition of water takes place according to anti-Markownikoff’s rule.
This is most stable as the repulsion between two methyl groups is least.
Among —CH 3 , —OCH 3 and CF 3 , CH 3 and —OCH 3 are electron donating groups.
Hence, they activate the benzene ring.
In these the order of activation is —OCH 3 —CH 3 while —CF 3 group deactivates the benzene nucleus.
So, it shows lower rate of electrophilic substitution on benzene ring.
Thus, order of electrophilic substitution is :
The staggered form of ethane has the following structure and the dihedral angle is 60 o , which means ‘H’ atoms are at an angle of 60 o to each other.
Metallic sodium does not react with 2-butyne because 2-butyne does not have acidic hydrogen.
The compound given is hex - 1 - en - 4 - yne Structure is a double bond has one bond and one sigma bond.
A triple bond has one bond and two bonds.
So, the number of bonds For sigma bonds, count all carbon carbon bonds and carbon hydrogen bonds except the bonds in double bond and triple bond.
The number of sigma bonds = 13 So, total number of bonds Answer : option (C) 13 and 3