[THC]

If 2-chloropropane reacts with SH-, will a thiol or an alkene be the resulting product?

And what if 2-chloropropane reacted with OH- instead, what would be the primary product? And how do I know?

[playstyles]

is this 1,1-dichloropropane; 1,2-dichloropropane, or 2,2-dicholorpropane?

[THC]

is this 1,1-dichloropropane; 1,2-dichloropropane, or 2,2-dicholorpropane?

Just 1-chloropropane and nothing else

[CopperSmurf]

Sounds like you would get a mixture of the thiol and alkene, but more of the thiol would be my best guess (by Sn1).

if 2-chloropropane reacted with OH- instead, you'd get 2-propanol (Sn1)

[macman104]

SH- is not a good base at all (well, I shouldn't say not a good base at all, but certainly not as good as OH-) and is more nucleophilic than the OH-.  I would expect only substitution from the thiol, and depending on what temperature and solvent used, possibly some elimination from the OH-.

[nj_bartel]

I'd expect pretty significant elimination with the OH^-, via E2, and mostly substitution via S_N2 for SH^-.  It's hard to say without reaction conditions though.

[macman104]

I'd expect pretty significant elimination with the OH^-, via E2, and mostly substitution via S_N2 for SH^-.  It's hard to say without reaction conditions though.

I agree about reaction conditions, especially for the borderline case of a secondary chloride.

[Doom91]

it depends on the solvent
for example dimethyl sulphoxide would be polar enough to dissolve the ionic compound  ,it is not protic ,therefore for OH- and SH- , a large percentage of the product would be Substitution products as the nucleophilicity of the chemical species would not be suppressed
 
however if use a polar protic solvent like water, the nucleophilicity of the chemical species OH- and SH- would be suppressed, therefore a majority of the products would be elimination products
       
If assuming the chemical species OH- and SH- are used in the same solvent,SH- would be a comparatively stronger conjugate base of H2S as compared to OH- as a conjugate base of water,this is because as oxygen is more electro-ve than sulphur , the dipole moment btw sulphur and hydrogen would be much smaller than the dipole moment btw oxygen and hydrogen, hence this would cause the -ve charge on S to be less than the -ve charge on O
Therefore,as SH- is more stable than Oh- , the eqm lies more to the right , hence less likely to act as the base, therefore there would be greater % of SN products as compared to the % of substitution products of OH-
    hope u will understand my explanation