Chem P1 doubts - Org Chem

[Lynn Gweeny]

Hello,

could somebody explain the attached problems? Also, for Q13, how can you infer the polarity with a rough idea of electronegativity? Do you need to know the values for some elements?

Thanks in advance!

[kw0573]

Without databooklets you just have to know 13. Rememeber that alkanes are non polar, but ketones, haloalkanes, and amines are all polar. 13. B

Electrophilic substitution adds something to the aromatic ring carbon, not on a substituent. 36. C

A lot of enantiomers contain a carbon bonded to 4 different groups. In 37. B, the third carbon from the left is bonded to an ethyl group, hydrogen, bromine, and a methyl group. 

[IB_taking_over]

13. 

The answer is C-H in the CH4

a. C=O : 3.4-2.6= .8 || b. C-H : 2.6-2.2= .4 || c. C-Cl : 3.2-2.6= .6 || d. N-H : 3.0-2.2= .8 (For reference; I am aware you don't get these values on Paper 1)

You can deduce based on the trend that C-H will be less polar than N-H and that C=O will be more polar than C-Cl.

So, now we are down to C-H and C-Cl.

We know that Cl has a relatively high electronegativity as it's on the far right. Which means that C-Cl has more polarity. (For the most part, if one of the options has a N, O, F, or Cl they are going to be the most polar option.)

So, C-H is the answer. 

[Lynn Gweeny]

4 hours ago, kw0573 said:

Without databooklets you just have to know 13. Rememeber that alkanes are non polar, but ketones, haloalkanes, and amines are all polar. 13. B

Electrophilic substitution adds something to the aromatic ring carbon, not on a substituent. 36. C

A lot of enantiomers contain a carbon bonded to 4 different groups. In 37. B, the third carbon from the left is bonded to an ethyl group, hydrogen, bromine, and a methyl group. 

Ohhh thank you!