Draw the organic product(s) formed when CH3CH2CH2OH is treated with each reagent. a. H2SO4 e. SOCl2, pyridine i. [1] TsCl, pyridine; [2] NaSH b. NaH f. PBr3 j. POCl3, pyridine c. HCl ZnCl2 g. TsCl, pyridine d. HBr h. [1] NaH; [2] CH3CH2Br

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Explanation:

In this case, we are treating an alcohol with different reagents to give different products. In the picture below you have the draw of each product, and here, the explanation.

a) H₂SO₄: In this case, it will promote an E1 reaction forming an alkene as product.

b) NaH: In this case, it will promote an acid base reaction, and the final product will be an alcoxide.

c) HCl/ZnCl₂: In this case, primary alcohols reacts with this reactive to form an alkyl halide.

d) HBr: Another reaction forming an alkyl halide with bromine in this case.

e) SOCl₂: Similar to c), it will form an alkyl chloride.

f) PBr₃: Similar to d, forming an alkyl bromide.

g) TsCl: Here it will be a sustitution reaction, where tosilate replaces the OH

h) NaH/CH₃CH₂Br: In this case it will form an alcoxide and then, a larger alkyl.

i) TsCl/NaSH: First it will form a tosilate and then, the SH group. A sustitution reaction.

j) POCl₃: A sustitution reaction forming an alkyl chloride.

See picture below for drawings.

Hope this helps

electron microscopy and scanning tunneling microscopes gives us a greater understanding of atoms by being able to see the atom itself, and lets us see some atomic reactions.

explanation:

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