how is this drawn?
 

propyne should be reacted with silver or cupric chloride so that it's acidic hydrogen gets displaced by silver or copper.react the product with 2-bromo propane.bromine being a good leaving group would combine with silver of propyne and you would get 2-yne,4-methyl pentyne.follow it by sodium ethanol process by which alcohol will be formed and react it with RED P+IODINE

How would this reaction be drawn?

Species involved in the reactions -

Propyne is a three carbon (prop-) alkyne (-yne; there's a triple bond between two carbons):

CH3-C-(triple bond)-CH

Sorry, I can't draw here.

Silver is a +1 cation, and chloride is a -1 anion. Cupric refers to the +2 cation of copper. These are some of the only chlorides which are not soluble in water, which may suggest that there is an exceptionally strong bond between them. Or, it may mean that silver and copper(II) ions are especially hydrophobic for some reason.

AgCl and CuCl2

2-bromopropane is a three carbon chain with bromine attached to the second (middle) carbon:

CH3-CHBr-CH3

I guess from context that the "silver of propyne" is

CH3-C-(triple bond)-CAg

I'm not sure if it's more of an ionic or covelant bond between C and Ag.

2-yne,4-methyl pentyne is a five carbon chain with a triple bond between the second and third carbons and a CH3 group on the fourth carbon. This name is incorrect by the way if I am not mistaken. The correct IUPAC name would be 4-methyl-2-pentyne, I think.

CH3-C-(triple bond)-C-CH(CH3)2

The "sodium ethanol process" may refer to the salt sodium ethoxide (I think that's how it's called; it might be sodium ethanoate too) in an ethanol solvent:

NaOCH2HC3 in EtOH

Red P is elemental phosphorus.

Reactions -

The first reaction using AgCl will look like this:

CH3-C-(triple bond)-CH + AgCl => CH-C-(triple bond)-CAg

This assumes that the silver ion will separate from the chloride ion and allow the reaction to occur. Terminal alkynes in general give insoluble salts with silver ion and other heavy metal cations.

Using cupric chloride to do this reaction seems impossible, as cupric ion has a +2 charge and can't just displace a hydrogen ion that has a +1 charge. Are you sure you didn't mean cuprous chloride?

If you have to show electron pushing arrows I don't know what to say. My textbook doesn't show that diagram, and it's not obvious to me how it would work.

The next reaction is probaby an SN2 mechanism; maybe it's SN1. I'm not sure. The nucleophile is the carbon in the "silver of propyne" that is adjacent to the silver atom. If it's an ionic bond between the two, then the silver will have already dropped off before the reaction, and the nucleophile will be a fully ionized propynide (is that the right name?) ion:

CH-C-(triple bond)-C:-

This SN2 (SN1?) reaction will end up with the "silver of propyne" being attached to the bromide, except that the silver and the bromine atoms will be lost from the molecule. This results in the 4-methyl-2-pentyne:

CH-C-(triple bond)-CAg + CH3-CHBr-CH3 => CH3-C-(triple bond)-C-CH(CH3)2 => CH3-C-(triple bond)-C-CH(CH3)2

When sodium ethanoate in ethanol is added, then there would be (if I have this right) a nucleophilic addition reaction of the ethoxide to the alkyne. I'm not sure which carbon the ethoxide ion will bond with though. My textbook says that it's a Markovnikov addition that forms the most stable anion between the two possible anionic alkenes. But in this case there are no hydrogens on either carbon and I don't know which would result in the more stable anion. It will be one of these two possibilities:

CH3C-(triple bond)-CCH(CH3)2 + (-)OCH2CH3 =>
CH3C(OCH2CH3)=C(-)CH(CH3)2

CH3C-(triple bond)-CCH(CH3)2 + (-)OCH2CH3 =>
CH3C(-)=C(OCH2CH3)CH(CH3)2

Unfortunately, no alcohol is formed by either of these reactions, so according to the description it must be something else. However, I don't know what else could be meant by a "sodium ethanol process".

The next step from these would be for the resulting anions to take a hydrogen from ethanol (I am abbreviating ethanol EtOH and ethoxide EtO(-))

CH3C(OCH2CH3)=C(-)CH(CH3)2 + EtOH => CH3C(OCH2CH3)=CHCH(CH3)2 + EtO(-)

CH3C(-)=C(OCH2CH3)CH(CH3)2 + EtOH => CH3CH=C(OCH2CH3)CH(CH3)2 + EtO(-)

Apparently we're supposed to have an alcohol at this point. I will call it ROH. When you heat an alcohol with red phosphorus and iodine, you get an alkyl iodide:

3 ROH + (3/2) I2 + P => 3 RI + H3PO4

If anyone knows the missing "sodium ethanol process" that results in an alcohol, please share with us!