Roth IRA excess contribution and withholding

[compoundinterest]

We made an excess ROTH Ira contribution for 2019.

This month, I withdrew our initial contribution ($6000). I made the mistake of electing for withholding on both federal and state taxes on the excess contribution earnings.

For myself:

Original contribution: $6000 (contributed spring 2019)
2019 Earnings withdrawl: $255.43 (withdrawn in 2020)
Federal withholding: $72.99
State withholding: $36.50

I think I really messed myself up with the withholding for turbotax. I was planning to do a "dummy" 2019 1099-R to pay tax on the earning as well as the penalty (and avoid an amended 2019 return). Now, as I *think* the withholding counts toward 2020 while the earnings are taxable in 2019, how should I handle this?  How do I address in turbotax?

Thanks in advance for any responses.

[dmertz]

You have figure this out correctly.  The $255.43 is taxable on your 2019 tax return but all of the withholding will be credited on your 2020 federal and state tax returns.

To handle this in TurboTax, the code JP 2020 Form 1099-R will need to be entered into both 2019 and 2020 TurboTax.  When you indicate to 2019 TurboTax that this is a 2020 Form 1099-R, 2019 TurboTax will include the $255.43 on Form 1040 line 4b and on Form 5329, and will ignore the tax withholding.  When you indicate to 2020 TurboTax that this is a 2020 Form 1099-R, 2020 TurboTax will ignore the taxable amount and credit you with the tax withholding.

As you mentioned, to report this on your 2019 tax return now instead of amending later, you'll need to enter a dummy code JP 2020 Form 1099-R.  In that dummy Form 1099-R I would omit the tax withholding just to ensure that 2019 TurboTax does not try to include the form details in your e-filing.  The explanation statement the should be present with your 2019 tax return will suffice.

[compoundinterest]

Thank you!