What is conditions for back bonding ?

CONDITIONS FOR BACK BONDING:

(1) Both of the atoms bonded with back Bonding are must be present in 2nd-2nd or 2nd-3rd period.

(2) One of the atoms has lone pair and another have vacant Orbital and direction of back Bonding depends upon vacant Orbital.

(3) The donor atom must have localized donatable electron pair. In general these are later half second period P - block elements (F, O, N and C).(4) The acceptor atom must have low energy empty orbital which generally are np or nd orbitals. Small and similar sized orbital’s favour overlap.

BCl3 ,BBr3 BI3 and B(Me)3 are electron deficient but do not undergo dimerization why ?

BCl_3 ,BBr₃ BI₃ and B(Me)₃ although  they are electron deficient compound but do not undergo dimerisation  because  of steric factor in demmeric formed.

Why some compounds are undergoes dimerization ?

Compound are dimerized due to following reasons:

(1) If there is no steric crowding and back bonding in a molecules then bridge bond formed and molecules dimerised and stabilized and dimerisation are more stable than back bonding.

(2) Most of the electron deficient compound attains stability by performing back bonding or they undergo dimerisation provided certain conditions are fulfilled

(3) BCl_3 ,BBr₃ BI₃ and B(Me)₃ although  they are electron deficient compound but do not undergo dimerisation  because  of steric factor in demmeric formed

How do you calculate the number of all possible structural isomers for alkanes?

Yes we can find all possible structural isomers Of a given hydrocarbon.

Number of total structural isomers = 2^(N-4)+1

Where N is number of carbons in hydrocarbon

Example: (1)
Number of structural somers of butane (C4H10) is 

SOLUTION:

Where , n=4
=2^(4-4)+1
=2^(0)+1
=1+1
=2

Example: (2)
Number of structural somers of butane (C5H12) is.

SOLUTION:

Where , n=5
=2^(5-4)+1
=2^(1)+1
=2+1
=3

Example: (3)
Number of structural somers of butane (C6H14) is.

SOLUTION:

Where , n=6
=2^(6-4)+1
=2^(2)+1
=4+1
=5

Example: (4)
Number of structural somers of butane (C7H16) is.

SOLUTION:

Where , n=7
=2^(7-4)+1
=2^(3)+1
=8+1
=9

Example: (5)
Number of structural somers of butane (C8H18) is.

SOLUTION:

Where , n=8
=2^(8-4)+1
=2^(4)+1
=16+1
=17

Example: (5)
Number of structural somers of butane (C8H18) is.

SOLUTION:

Where , n=9
=2^(9-4)+1
=2^(5)+1
=32+1
=33

How to find number of stereoisomers in a large compound (polyenes) containing double bonds?

Number of stereoisomer of polyenes can calculated by following formulas .