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The reaction is such:

I've counted the number of carbons on each side: LHS=4, RHS=8 My working is such:

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    $\begingroup$ The reaction is a self aldol condensation. The enolate of the aldehyde attacks a molecule of un-enolised aldehyde. The resulting intermediate eventually loses water to give your product, as drawn. Give the curly arrows a try and post what you end up with :) $\endgroup$ – NotEvans. Dec 30 '15 at 23:57
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    $\begingroup$ You do see that your last intermediate would not arrive at your product, don’t you? $\endgroup$ – Jan Dec 31 '15 at 0:30
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Conceptually, we can think of this reaction as occurring in three stages. I've provided diagrams for the self-condensation of acetaldehyde, however you should be able to draw an analogous mechanism for the self-condensation of butanal (your aldehyde).

Your mechanism is essentially correct, except for the final elimination step, which as you've drawn it gives the wrong product.

1. Enolisation of the aldehyde

enter image description here
Source: Organic Chemistry, 2 ed., Warren, Clayden, Wothers & Greeves

As you guessed, the hydroxide is basic enough to cause enolisation of the aldehyde. Using KOH/MeOH, we form the enolate under thermodynamic control, essentially meaning the enolate is in equilibrium with the aldehyde.

2. Addition of the enolate into another molecule of aldehyde

enter image description here
Source: Organic Chemistry, 2 ed., Warren, Clayden, Wothers & Greeves

The newly formed enolate may then attack an un-enolised molecule of the aldehyde (this is okay, since we've established that the aldehyde and the enolate are in equilibrium under these conditions). This nucleophilic attack leads to a beta-alkoxy aldehyde (the structure on the right), which will quickly pick up a proton to form a beta-hydroxy aldehyde.

Under these basic conditions (KOH, MeOH), we may be able to isolate the hydroxyaldehyde, however, they are prone to elimination (see below), which is what must happen to form the product you require.

3. Elimination of hydroxide

enter image description here
Source: Organic Chemistry, 2 ed., Warren, Clayden, Wothers & Greeves

The beta-hydroxyaldehyde formed (see above) is itself able to undergo enolisation, to form a new enolate. Rather than attacking another molecule of aldehyde intermolecularly, it may instead do an intramolecular E1cb type elimination, kicking out hydroxide (which is okay under these basic conditions), and giving you the product as drawn.

Your elimination step isn't the most favourable pathway to get rid of that alcohol, and infact the elimination only occurs under these conditions due to assistance form the carbonyl. The quote below puts this rather well.

You cannot normally eliminate water from an alcohol in basic solution as hydroxide is a bad leaving group. It is the carbonyl group that allows elimination here: these are E1cB reactions.
Source: Organic Chemistry, 2 ed., Warren, Clayden, Wothers & Greeves

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In your last elimination step the OH- should attack the hydrogen on the carbon between the alcohol and the ketone.

This then forms an enol, which then reverts to the ketone. This pushes the pi-bond across one carbon. The alcohol then leaves.

You should end up with almost the same molecule you've put down, but the alkene should be one to the left.

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