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New answer posted
11 months agoContributor-Level 10
(a) Propylbenzene (b) 3-Methylpentanenitrite (c) 2, 5-Dimethylheptane
(d) 3-Bromo- 3-chloroheptane (e) 3-Chloropropanal (f) 2, 2-Dichloroethanol
New question posted
11 months agoNew answer posted
11 months agoContributor-Level 10
11.13
2,2,4-Trimethylpentan-3-ol
The naming of the compound usually starts with numbering the carbons in the chain. The lower set of locants are chosen for this purpose, while in this case numbering under this condition is done from the left side. Once the carbons are mentioned, the position of -OH group is numbered and -ol is added as the suffix.
2. 5-Ethylheptane-2,4-diol
Here the longest chain is the straight chain. For such situations, numbering should be such that the functional groups should be denoted by the smallest number. Ethyl group is named at the beginning as it's a side chain.
3. Butane-2,3-diol
In this system, gly
New question posted
11 months agoNew answer posted
11 months agoContributor-Level 10
Single bond contains only one sigma bond and double bond contains one sigma and one pi bond.

- In this structure, nine single bonds and three double bonds are present. So, there are 12σ and 3π bonds present.
- In this structure, eighteen single bonds are present. So, there are only 18σ bonds present.
- In this structure, four single bonds are present. So, there are only 4σ bonds present.
- In this structure, four single bonds and two double bonds are present. So, there are 6σ and 2π bonds present.
- In this structure, five single bonds and one double bond are present. So, there are 6σ and 1π bonds present.
- In this structure, seven single bonds
New answer posted
11 months agoContributor-Level 10
4.26. In BF3, B atom is sp2 hybridised. In NH3, N is sp3 hybridised.
After the reaction, hybridisation of B changes from sp2 to sp3, while the hybridisation of N remains the same.
New answer posted
11 months agoContributor-Level 10
When N –propyl methyl ether reacts with HBr, it forms propanol and bromomethane, n- propyl methyl ether will cleave at O . and H+ will attack at O, and Br- will attack CH +
2. When Ethoxybenzene Reacts with HBr, it forms Phenol and bromoethane, Ethoxybenzene cleaves at H+ will attack at O, and Br- will attack
3. When nitrating mixture reacts with ethoxy benzene introduction of nitro group is occurred at para position as it will give the stable product without
4. As HI is a strong nucleophile it will protonate the oxygen, to form a good leaving And I- will attack at C (CH3)3+ to give tert- butyl iodide and ethanol
New question posted
11 months agoNew answer posted
11 months agoContributor-Level 10
4.25. Electronic configuration of 13Al in ground state= 1s2 2s2 2p6 3s2 3p1
Electronic configuration of 13Al in excited state = 1s2 2s2 2p6 3s1 3px13py1
Hence, hybridisation will be sp2 and this makes the geometry to be Trigonal planar (in case of AlCl3).
In AlCl–4, the empty 3pz orbital is also involved. So, the hybridisation is sp3 and the shape is tetrahedral
New answer posted
11 months agoContributor-Level 10
11.11. Set (ii) is appropriate Because CH3Br is only a nucleophile whereas CH3ONa is nucleophile as well as strong base, so the elimination reaction can occur
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